NCERT Solutions for Class 10 Maths Chapter 6: Triangles - Exercise 6.3

Exercise 6.3

1. State which pairs of triangles in the following figure are similar? Write the similarity criterion used by you for answering the question and also write the pairs of similar triangles in the symbolic form:

(image will be uploaded soon)

Ans: 

1. In $\triangle \mathrm{ABC}$ and $\triangle \mathrm{PQR}$

$\angle \mathrm{A}=\angle \mathrm{P}$

$\angle \mathrm{B}=\angle \mathrm{Q}$

$\angle \mathrm{C}=\angle \mathrm{R}$

$\therefore$ By AAA criterion of similarity, $\triangle \mathrm{ABC} \sim \triangle \mathrm{PQR}$

2. In $\triangle \mathrm{ABC}$ and $\triangle \mathrm{QRP}$

$\frac{\mathrm{AB}}{\mathrm{QR}}=\frac{\mathrm{BC}}{\mathrm{RP}}=\frac{\mathrm{AC}}{\mathrm{QP}}=\frac{1}{2}$

$\therefore$ By SSS criterion of similarity, $\triangle \mathrm{ABC} \sim \triangle \mathrm{QRP}$

3. In $\triangle \mathrm{LMP}$ and $\triangle \mathrm{DEF}$

$\frac{\mathrm{LM}}{\mathrm{DE}}=\frac{2.7}{4}, \frac{\mathrm{LP}}{\mathrm{DF}}=\frac{1}{2}$

The sides are not in the equal ratios, Hence the two triangles are not similar.

4. In $\triangle \mathrm{MNL}$ and $\triangle \mathrm{QPR}$

$\angle \mathrm{M}=\angle \mathrm{Q}$

$\frac{\mathrm{MN}}{\mathrm{QP}}=\frac{\mathrm{ML}}{\mathrm{QR}}=\frac{1}{2}$

$\therefore$ By SAS criterion of similarity, $\triangle \mathrm{MNL} \sim \triangle \mathrm{QP} \mathrm{R}$

5. In $\triangle \mathrm{ABC}$ and $\triangle \mathrm{EFD}$

$\angle \mathrm{A}=\angle \mathrm{F}, $

$\frac{AB}{FD}=\frac{BC}{FD}=\frac{1}{2}$

$\therefore$ By SAS criterion of similarity, $\triangle \mathrm{ABC} \sim \triangle \mathrm{EFD}$

6. In $\triangle \mathrm{DEF}$ and $\triangle \mathrm{PQR}$

Since, sum of angles of a triangle is $180^{\circ}$, Hence, $\angle \mathrm{F}=30^{\circ}$ and $\angle \mathrm{P}=70^{\circ}$

$\angle \mathrm{D} =\angle \mathrm{P}$

$\angle \mathrm{E} =\angle \mathrm{Q}$

$\angle \mathrm{F} =\angle \mathrm{R}$

$\therefore$ By AAA criterion of similarity, $\triangle \mathrm{DEF} \sim \triangle \mathrm{PQR}$

2. In the following figure, $\Delta \mathrm{ODC} \sim \Delta \mathrm{OBA}, \angle \mathrm{BOC}=125^{\circ}$ and $\angle \mathrm{CDO}=70^{\circ}$. Find $\angle \mathrm{DOC}, \angle \mathrm{DCO}$ and $\angle \mathrm{OAB}$

(image will be uploaded soon)

Ans: Given:

$\triangle \mathrm{ODC} \sim \triangle \mathrm{OBA}$

$\angle \mathrm{BOC}=125^{\circ}$

$\angle \mathrm{CDO}=70^{\circ}$

To find: $\angle D O C, \angle D C O$ and $\angle O A B$

Sol: Here, $B D$ is a line,

So, we can apply a linear pair on it.

$\angle B O C+\angle D O C=180^{\circ} (Linear Pair)$

$125^{\circ}+\angle D O C=180^{\circ}$

$\angle D O C=180^{\circ}-125^{\circ}$

$\angle D O C=180^{\circ}-125^{\circ}$

$\angle D O C=55^{\circ}$

Now in $\triangle \mathrm{DCO}$

$\angle C D O+\angle D C O+\angle D O C=180^{\circ}$

$70^{\circ}+\angle D C O+55^{\circ}=180^{\circ}$

$125^{\circ}+\angle D C O=180^{\circ}$

$\angle D C O=180^{\circ}-125^{\circ}$

$\angle D C O=55^{\circ}$

Now it is given that

$\triangle O D C \sim \triangle O B A$

Hence,

$\angle \mathrm{DCO}=\angle \mathrm{OAB}$

$55^{\circ}=\angle \mathrm{OAB}$

$\angle \mathrm{OAB}=55^{\circ}$

Now in $\triangle \mathrm{DCO}$

$\angle \mathrm{CDO}+\angle \mathrm{DCO}+\angle \mathrm{DOC}=180^{\circ} \quad$ (Sum of all angles of triangle is $180^0$ 

$70^{\circ}+\angle \mathrm{DCO}+55^{\circ}=180^{\circ}$ 

$125^{\circ}+\angle \mathrm{DCO}=180^{\circ}$ 

$\angle \mathrm{DCO}=180^{\circ}-125^{\circ}$ $\angle \mathrm{DCO}=55^{\circ}$

Now, it is given that

$\Delta \mathrm{ODC} \sim \triangle \mathrm{OBA}$

Hence.

$\angle D C O=\angle O A B$(Corresponding angles of a similar triangles are equal)

$55^{\circ}=\angle O A B$

3. Diagonals AC and BD of a trapezium ABCD with AB \|DC intersect each other at the point $\mathrm{O}$. Using a similarity criterion for two triangles, show that $\frac{\mathrm{AO}}{\mathrm{OC}}=\frac{\mathrm{OB}}{\mathrm{OD}}$

Ans: In $\Delta \mathrm{DOC}$ and $\triangle \mathrm{BOA}$

$\angle C D O=\angle A B O$ (Alternate interior angles as $A B \| C D)$

$\angle \mathrm{DCO}=\angle \mathrm{BAO}$ (Alternate interior angles as $\mathrm{AB} \| \mathrm{CD})$

$\angle \mathrm{DOC}=\angle \mathrm{BOA}$ (Vertically opposite angles $)$

$\therefore \Delta \mathrm{DOC} \sim \Delta \mathrm{BOA}$ (AAA similarity criterion)

$\therefore \frac{\mathrm{DO}}{\mathrm{BO}}=\frac{\mathrm{OC}}{\mathrm{OA}} \quad($ Corresponding sides are proportional)

$\Rightarrow \frac{\mathrm{OA}}{\mathrm{OC}}=\frac{\mathrm{OB}}{\mathrm{OD}}$

4.  In the figure, $\frac{\mathrm{QR}}{\mathrm{QS}}=\frac{\mathrm{QT}}{\mathrm{PR}} \text { and } \angle 1=\angle 2 \text {.Show that } \Delta \mathrm{PQS} \sim \Delta \mathrm{TQR}$

(image will be uploaded soon)

Ans: In $\delta \mathrm{PQR}, \angle \mathrm{PQR}=\angle \mathrm{PRQ}$

$\therefore \mathrm{PQ}=\mathrm{PR}(\mathrm{i})$

Given 

$\frac{\mathrm{QR}}{\mathrm{QS}}=\frac{\mathrm{QT}}{\mathrm{PR}}$

Using (i), we obtain

(image will be uploaded soon)

$\frac{\mathrm{QR}}{\mathrm{QS}}=\frac{\mathrm{QT}}{\mathrm{QP}}$

In $\triangle \mathrm{PQS}$ and $\triangle \mathrm{TQR}$,

$\frac{\mathrm{QR}}{\mathrm{QS}}=\frac{\mathrm{QT}}{\mathrm{QP}}[\text { [Using (ii) }]$

$\angle \mathrm{Q}=\angle \mathrm{Q}$

$\therefore \Delta \mathrm{PQS} \sim \Delta \mathrm{TQR} \quad[\text { SAS similarity criterion }]$

5. $\mathrm{S}$ and $\mathrm{T}$ are point on sides $\mathrm{PR}$ and $\mathrm{QR}$ of $\triangle \mathrm{PQR}$ such that $\angle \mathrm{P}=\angle \mathrm{RTS}$. Show that $\triangle \mathrm{RPQ} \sim \Delta \mathrm{RTS}$.

Ans: Given: $\Delta P Q R$

and the points $S$ and $T$ on sides PR and QR.

Such that $\angle P=\angle R T S$

(image will be uploaded soon)

To Prove: $\triangle \mathrm{RPQ} \sim \Delta$ RTS.

Proof:

In $\triangle \mathrm{RPQ}$ and $\triangle \mathrm{RTS}$.

$\angle P=\angle R T S$

(Given)

And $\angle \mathrm{PRQ}=\angle \mathrm{TRS}=\angle \mathrm{R}$

(Common)

So, $\triangle \mathrm{RPQ} \sim \Delta \mathrm{RTS}$.

(AA similarity)

Hence proved

6. In the following figure, if $\triangle \mathrm{ABE} \cong \triangle \mathrm{ACD}$, show that $\triangle \mathrm{ADE} \sim \Delta \mathrm{ABC}$.

Ans: 

(image will be uploaded soon)

We know that the corresponding portions of two triangles that are congruent to each other are equal.

The two triangles are comparable if one of their angles is equal to one of the other triangle's angles, and the sides that include these angles are proportionate.

For two triangles, this is known as the SAS (Side - Angle - Side) similarity criteria.

In $\triangle \mathrm{ABE}$ and $\triangle \mathrm{ACD}$

$\mathrm{AD}=\mathrm{AE}(\triangle \mathrm{ABE} \cong \Delta \mathrm{ACD} \text { given }) \ldots \ldots \ldots \text { (1) }$

$\mathrm{AB}=\mathrm{AC}(\triangle \mathrm{ABE} \cong \triangle \mathrm{ACD} \text { given })$

Now Consider $\triangle A D E$ and $\triangle A B C$

and $\angle$ DAE $=\angle B A C$ (Common angle)

Thus, $\triangle$ ADE $\sim A$ ABC (SAS criterion)

7. In the following figure, altitudes $\mathrm{AD}$ and $\mathrm{CE}$ of $\Delta \mathrm{ABC}$ intersect each other at the point, P. Show, that:

(image will be uploaded soon)

1. $\triangle \mathrm{AEP} \sim \Delta \mathrm{CDP}$

Ans: When two angles from one triangle are equivalent to two angles from another triangle, the two triangles are said to be comparable.

For two triangles, this is known as the AA similarity criteria.

In $\triangle \mathrm{AEP}$ and $\triangle \mathrm{CDP}$

$[\because \mathrm{CE} \perp \mathrm{AB}$ and $\mathrm{AD} \perp \mathrm{BC} ;$ altitudes $]$

$\angle A P E=\angle C P D$ (Vertically opposite angles)

$\Rightarrow \triangle$ AEP $\sim \triangle$ CPD (AA criterion)

2. $\triangle \mathrm{ABD} \sim \Delta \mathrm{CBE}$

Ans: When two angles from one triangle are equivalent to two angles from another triangle, the two triangles are said to be comparable.

For two triangles, this is known as the AA similarity criteria.

In $\triangle \mathrm{ABD}$ and $\triangle \mathrm{CBE}$

$\angle \mathrm{ADB}=\angle C E B=90^{\circ}$

$\angle \mathrm{ABD}=\angle C B E \text { (Common angle) }$

$\Rightarrow \triangle \mathrm{ABD} \sim \Delta C B E \text { (AA criterion) }$

3. $\triangle \mathrm{AEP} \sim \triangle \mathrm{ADB}$

Ans: When two angles from one triangle are equivalent to two angles from another triangle, the two triangles are said to be comparable.

For two triangles, this is known as the AA similarity criteria.

In $\triangle \mathrm{AEP}$ and $\triangle \mathrm{ADB}$

$\angle \mathrm{AEP}=\angle \mathrm{ADB}=9 \mathrm{O}^{\circ}$

$\angle \mathrm{PAE}=\angle \mathrm{BAD} \text { (Common angle) }$

$\Rightarrow \triangle \mathrm{AEP} \sim \triangle \mathrm{ADB} \text { (AA criterion) }$

4. $\Delta \mathrm{PDC} \sim \Delta \mathrm{BEC}$

Ans: When two angles from one triangle are equivalent to two angles from another triangle, the two triangles are said to be comparable.

For two triangles, this is known as the AA similarity criteria.

In $\triangle \mathrm{PDC}$ and $\triangle \mathrm{BEC}$

$\angle \mathrm{PDC}=\angle \mathrm{BEC}=9 \mathrm{O}^{\circ}$

$\angle \mathrm{PCD}=\angle \mathrm{BCE} \text { (Common angle) }$

$\Rightarrow \triangle \text { PDC } \sim \triangle \mathrm{BEC} \text { (AA criterion)}$

8. $\mathrm{E}$ is a point on the side AD produced of a parallelogram $\mathrm{ABCD}$ and $\mathrm{BE}$ intersects $\mathrm{CD}$ at $\mathrm{F}$. Show that $\triangle \mathrm{ABE} \sim \Delta \mathrm{CFB}$

(image will be uploaded soon)

Ans:

In $\triangle \mathrm{ABE}$ and $\triangle \mathrm{CFB}$,

$\angle \mathrm{A}=\angle \mathrm{C}$ (Opposite angles of a parallelogram)

$\angle \mathrm{AEB}=\angle \mathrm{CBF}$ (Alternate interior angles as $\mathrm{AE} \| \mathrm{BC})$

$\therefore \Delta \mathrm{ABE} \sim \Delta \mathrm{CFB}$ (By AA similarity criterion)

9. In the following figure, $\mathrm{ABC}$ and AMP are two right triangles, right angled at B and M respectively, prove that:

(image will be uploaded soon)

1. $\Delta \mathrm{ABC} \sim \Delta \mathrm{AMP}$

2. $\frac{\mathrm{CA}}{\mathrm{PA}}=\frac{\mathrm{BC}}{\mathrm{MP}}$

Ans: In $\triangle \mathrm{ABC}$ and $\triangle \mathrm{AMP}$

$\angle \mathrm{ABC}=\angle \mathrm{AMP}\left(\operatorname{Each} 90^{\circ}\right)$ $\angle \mathrm{A}=\angle \mathrm{A}(\mathrm{Common})$

$\therefore \Delta \mathrm{ABC} \sim \Delta \mathrm{AMP}$ (By AA similarity criterion)

$\Rightarrow \frac{\mathrm{CA}}{\mathrm{PA}}=\frac{\mathrm{BC}}{\mathrm{MP}}$ (Corresponding sides of similar triangles are proportional)

10. $\mathrm{CD}$ and $\mathrm{GH}$ are respectively the bisectors of $\angle \mathrm{ACB}$ and $\angle \mathrm{EGF}$ such that $\mathrm{D}$ and $\mathrm{H}$ lie on sides $\mathrm{AB}$ and $\mathrm{FE}$ of $\triangle \mathrm{ABC}$ and $\triangle \mathrm{EFG}$ respectively. If $\triangle \mathrm{ABC} \sim$ $\Delta \mathrm{FEG}$, Show that:

(image will be uploaded soon)

(i) $\frac{\mathrm{CD}}{\mathrm{GH}}=\frac{\mathrm{AC}}{\mathrm{FG}}$

Ans: $\text { It is given that } \triangle \mathrm{ABC} \sim \Delta \mathrm{FEG}.$

$\therefore \angle \mathrm{A}=\angle \mathrm{F}, \angle \mathrm{B}=\angle \mathrm{E}, \text { and } \angle \mathrm{ACB}=\angle \mathrm{FGE}$

$\text { Since, } \angle \mathrm{ACB}=\angle \mathrm{FGE} $

$\therefore \angle \mathrm{ACD}=\angle \mathrm{FGH} \text { (Angle bisector) } $

$\text { And, } \angle \mathrm{DCB}=\angle \mathrm{HGE} \text { (Angle bisector) } $

$\text { In } \triangle \mathrm{ACD} \text { and } \Delta \mathrm{FGH} \text {, }$

$\angle \mathrm{A}=\angle \mathrm{F} \text { (Proved above) }$

$\angle \mathrm{ACD}=\angle \mathrm{FGH} \text { (Proved above) }$

$\therefore \Delta \mathrm{ACD} \sim \Delta \mathrm{FGH} \text { (By AA similarity criterion) }$

$\Rightarrow \frac{\mathrm{CD}}{\mathrm{GH}}=\frac{\mathrm{AC}}{\mathrm{FG}}$

(ii) $\triangle \mathrm{DCB} \sim \Delta \mathrm{HGE}$

Ans: $\text { In } \triangle \mathrm{DCB} \text { and } \triangle \mathrm{HGE} \text {, } $

$\angle \mathrm{DCB}=\angle \mathrm{HGE} \text { (Proved above) } $

$\angle \mathrm{B}=\angle \mathrm{E} \text { (Proved above) } $

$\therefore \triangle \mathrm{DCB} \sim \triangle \mathrm{HGE} \text { (By AA similarity criterion) }$

(iii) $\triangle \mathrm{DCA} \sim \Delta \mathrm{HGF}$

Ans: In $\Delta \mathrm{DCA}$ and $\Delta \mathrm{HGF}$,

$\angle \mathrm{ACD}=\angle \mathrm{FGH}$ (Proved above)

$\angle A=\angle F$ (Proved above)

$\therefore \Delta \mathrm{DCA} \sim \Delta \mathrm{HGF}$ (By AA similarity criterion)

11. In the following figure, $\mathrm{E}$ is a point on side CB produced of an isosceles triangle $\mathrm{ABC}$ with $\mathrm{AB}=\mathrm{AC}$. If $\mathrm{AD} \perp \mathrm{BC}$ and $\mathrm{EF} \perp \mathrm{AC}$, prove that $\triangle \mathrm{ABD} \sim$ $\triangle \mathrm{ECF}$

(image will be uploaded soon)

Ans: It is given that $\mathrm{ABC}$ is an isosceles triangle.

$\therefore \mathrm{AB}=\mathrm{AC}$

$\Rightarrow \angle \mathrm{ABD}=\angle \mathrm{ECF}$

In $\Delta \mathrm{ABD}$ and $\triangle \mathrm{ECF}$

$\angle \mathrm{ADB}=\angle \mathrm{EFC}\left(\operatorname{Each} 90^{\circ}\right)$

$\angle \mathrm{BAD}=\angle \mathrm{CEF}$ (Proved above)

$\therefore \Delta \mathrm{ABD} \sim \triangle \mathrm{ECF}$ (By using AA similarity criterion)

12. Sides $\mathrm{AB}$ and $\mathrm{BC}$ and median AD of a triangle $\mathrm{ABC}$ are respectively proportional to sides PQ and QR and median PM of $\triangle \mathrm{PQR}$ (see the given figure). Show that $\triangle \mathrm{ABC} \sim \triangle \mathrm{PQR}$.

(image will be uploaded soon)

Ans: Median equally divides the opposite side.

$\therefore \mathrm{BD}=\frac{\mathrm{BC}}{2}$ and $\mathrm{QM}=\frac{\mathrm{QR}}{2}$

Given that,

$\frac{\mathrm{AB}}{\mathrm{PQ}}=\frac{\mathrm{BC}}{\mathrm{QR}}=\frac{\mathrm{AD}}{\mathrm{PM}}$

$\Rightarrow \frac{\mathrm{AB}}{\mathrm{PQ}}=\frac{\frac{1}{2} \mathrm{BC}}{\frac{1}{2} \mathrm{QR}}=\frac{\mathrm{AD}}{\mathrm{PM}}$

$\Rightarrow \frac{\mathrm{AB}}{\mathrm{PQ}}=\frac{\mathrm{BD}}{\mathrm{QM}}=\frac{\mathrm{AD}}{\mathrm{PM}}$

In $\triangle \mathrm{ABD}$ and $\triangle \mathrm{PQM}$,

$\frac{\mathrm{AB}}{\mathrm{PQ}}=\frac{\mathrm{BD}}{\mathrm{QM}}=\frac{\mathrm{AD}}{\mathrm{PM}} \text { (Proved above) }$

$\therefore \Delta \mathrm{ABD} \sim \Delta \mathrm{PQM}$ (By SSS similarity criterion)

$\Rightarrow \angle \mathrm{ABD}=\angle \mathrm{PQM}$ (Corresponding angles of similar triangles)

In $\triangle \mathrm{ABC}$ and $\Delta \mathrm{PQR}$,

$\angle \mathrm{ABD}=\angle \mathrm{PQM} \text { (Proved above) }$

$\frac{\mathrm{AB}}{\mathrm{PQ}}=\frac{\mathrm{BC}}{\mathrm{QR}}$

$\therefore \Delta \mathrm{ABC} \sim \Delta \mathrm{PQR}$ (By SAS similarity criterion)

13. $\mathrm{D}$ is a point on the side $\mathrm{BC}$ of a triangle $\mathrm{ABC}$ such that $\angle \mathrm{ADC}=\angle \mathrm{BAC}$. Show that $\mathrm{CA}^{2}= \mathrm{CB.CD}$

(image will be uploaded soon)

Ans: $\text { In } \triangle \mathrm{ADC} \text { and } \triangle \mathrm{BAC} \text {, }$

$\angle \mathrm{ADC}=\angle \mathrm{BAC}$ (Given) $\angle \mathrm{ACD}=\angle \mathrm{BCA}$ (Common angle)

$\therefore \Delta \mathrm{ADC} \sim \triangle \mathrm{BAC}$ (By AA similarity criterion)

We know that corresponding sides of similar triangles are in proportion. $\therefore \frac{\mathrm{CA}}{\mathrm{CB}}=\frac{\mathrm{CD}}{\mathrm{CA}}$ $\Rightarrow \mathrm{CA}^{2}=\mathrm{CB} \cdot \mathrm{CD}$

14. Sides $\mathrm{AB}$ and AC and median AD of a triangle $\mathrm{ABC}$ are respectively proportional to sides PQ and PR and median PM of another triangle PQR. Show that $\triangle \mathrm{ABC} \sim \Delta \mathrm{PQR}$

(image will be uploaded soon)

Ans: Given that,

$\frac{\mathrm{AB}}{\mathrm{PQ}}=\frac{\mathrm{AC}}{\mathrm{PR}}=\frac{\mathrm{AD}}{\mathrm{PM}}$

Let us extend AD and PM up to point $E$ and $L$ respectively, such that $A D=$ DE and PM $=$ ML. 

Then, join $B$ to $E, C$ to $E, Q$ to $L$, and $R$ to $L .$

(image will be uploaded soon)

We know that medians divide opposite sides.

Therefore, $\mathrm{BD}=\mathrm{DC}$ and $\mathrm{QM}=\mathrm{MR}$

Also, $\mathrm{AD}=\mathrm{DE}$ (By construction)

And, $\mathrm{PM}=\mathrm{ML}$ (By construction)

In quadrilateral ABEC, diagonals AE and BC bisect each other at point D. Therefore, quadrilateral ABEC is a parallelogram.

$\therefore \mathrm{AC}=\mathrm{BE}$ and $\mathrm{AB}=\mathrm{EC}$ (Opposite sides of a parallelogram are equal) Similarly, we can prove that quadrilateral PQLR is a parallelogram and PR = QL, $P Q=L R$

It was given that

$\frac{\mathrm{AB}}{\mathrm{PQ}}=\frac{\mathrm{AC}}{\mathrm{PR}}=\frac{\mathrm{AD}}{\mathrm{PM}}$

$\Rightarrow \frac{\mathrm{AB}}{\mathrm{PQ}}=\frac{\mathrm{BE}}{\mathrm{QL}}=\frac{2 \mathrm{AD}}{2 \mathrm{PM}}$

$\Rightarrow \frac{\mathrm{AB}}{\mathrm{PQ}}=\frac{\mathrm{BE}}{\mathrm{QL}}=\frac{\mathrm{AE}}{\mathrm{PL}}$

$\therefore \triangle \mathrm{ABE} \sim \triangle \mathrm{PQL}$ (By SSS similarity criterion)

We know that corresponding angles of similar triangles are equal.

$\therefore \angle \mathrm{BAE}=\angle \mathrm{QPL} \ldots$ (1)

Similarly, it can be proved that $\triangle \mathrm{AEC} \sim \triangle \mathrm{PLR}$ and

$\angle \mathrm{CAE}=\angle \mathrm{RPL} \ldots$ (2)

Adding equation (1) and (2), we obtain

$\angle \mathrm{BAE}+\angle \mathrm{CAE}=\angle \mathrm{QPL}+\angle \mathrm{RPL}$

$\Rightarrow \angle \mathrm{CAB}=\angle \mathrm{RPQ} \ldots$ (3)

In $\triangle \mathrm{ABC}$ and $\triangle \mathrm{PQR}$,

$\frac{\mathrm{AB}}{\mathrm{PQ}}=\frac{\mathrm{AC}}{\mathrm{PR}}$ (Given)

$\angle \mathrm{CAB}=\angle \mathrm{RPQ}[\mathrm{Using}$ equation $(3)]$

$\therefore \triangle \mathrm{ABC} \sim \triangle \mathrm{PQR}$ (By SAS similarity criterion)

15. A vertical pole of a length $6 \mathrm{~m}$ casts a shadow $4 \mathrm{~m}$ long on the ground and at the same time a tower casts a shadow $28 \mathrm{~m}$ long. Find the height of the tower.

(image will be uploaded soon)

Ans: Let $\mathrm{AB}$ and $\mathrm{CD}$ be a tower and a pole respectively.

Let the shadow of $\mathrm{BE}$ and DF be the shadow of $\mathrm{AB}$ and $\mathrm{CD}$ respectively.

At the same time, the light rays from the sun will fall on the tower and the pole at the same angle.

Therefore, $\angle \mathrm{DCF}=\angle \mathrm{BAE}$

And, $\angle \mathrm{DFC}=\angle \mathrm{BEA}$

$\angle \mathrm{CDF}=\angle \mathrm{ABE}$ (Tower and pole are vertical to the ground)

$\therefore \Delta \mathrm{ABE} \sim \Delta \mathrm{CDF}$ (AAA similarity criterion)

$\Rightarrow \frac{\mathrm{AB}}{\mathrm{CD}}=\frac{\mathrm{BE}}{\mathrm{DF}}$

$\Rightarrow \frac{\mathrm{AB}}{6 \mathrm{~cm}}=\frac{28}{4}$

$\Rightarrow \mathrm{AB}=42 \mathrm{~m}$

Therefore, the height of the tower will be 42 metres.

16. If $\mathrm{AD}$ and $\mathrm{PM}$ are medians of triangles $\mathrm{ABC}$ and $\mathrm{PQR}$, respectively where $\Delta \mathrm{ABC} \sim \Delta \mathrm{PQR}$ Prove that $\frac{\mathrm{AB}}{\mathrm{PQ}}=\frac{\mathrm{AD}}{\mathrm{PM}}$

(image will be uploaded soon)

Ans: It is given that $\triangle \mathrm{ABC} \sim \Delta \mathrm{PQR}$

We know that the corresponding sides of similar triangles are in proportion. 

$\therefore \frac{\mathrm{AB}}{\mathrm{PQ}}=\frac{\mathrm{AC}}{\mathrm{PR}}=\frac{\mathrm{BC}}{\mathrm{QR}} \ldots(1)$

Also, $\angle \mathrm{A}=\angle \mathrm{P}, \angle \mathrm{B}=\angle \mathrm{Q}, \angle \mathrm{C}=\angle \mathrm{R} \quad \ldots$ (2)

Since AD and PM are medians, they will divide their opposite sides. $\therefore \mathrm{BD}=\frac{\mathrm{BC}}{2}$ and $\mathrm{QM}=\frac{\mathrm{QR}}{2}$

From equations (1) and (3), we obtain

$\frac{\mathrm{AB}}{\mathrm{PQ}}=\frac{\mathrm{BD}}{\mathrm{QM}}$

In $\triangle \mathrm{ABD}$ and $\triangle \mathrm{PQM}$,

$\angle B=\angle Q$ (Using equation (2))

$\frac{\mathrm{AB}}{\mathrm{PQ}}=\frac{\mathrm{BD}}{\mathrm{QM}}$

(Using equation (4))

$\therefore \triangle \mathrm{ABD} \sim \triangle \mathrm{PQM}$ (By SAS similarity criterion)

$\Rightarrow \frac{\mathrm{AB}}{\mathrm{PQ}}=\frac{\mathrm{BD}}{\mathrm{QM}}=\frac{\mathrm{AD}}{\mathrm{PM}}$

CBSE Class 10 Maths Chapter 6 Exercise 6.3 Solutions Free PDF

If you want to do a quick revision of various tips and tricks explained in NCERT Solutions for Class 10 Maths ex 6.3 you can download its PDF format from the official website of Vedantu. The NCERT Class 10 Exercise 6.3 Solutions PDF is readily available and can be downloaded or printed for ease of use. Once you have saved the PDF of Exercise 6.3 Class 10 NCERT Solutions on your system, you do not need to bother about internet connection during exam time preparations.

Topic Covered in Class 10 Maths Chapter 6 Exercise 6.3

Exercise 6.3 of Class 10 NCERT Maths is based on the topic ‘criteria for similarity of triangles.’

Criteria for Similarity of Triangles: Two triangles are said to be similar if their corresponding angles are equal and their corresponding sides are in the same proportion.

Different Theorems Covered in Exercise 6.3

Theorem 1: If in two triangles, corresponding angles are equal then their corresponding sides are in the same ratio or proportion. Hence, two triangles are said to be similar.

Theorem 2: If in two triangles, sides of one triangle are proportional to the sides of another triangle, then their corresponding angles are equal. Hence, the two triangles are said to be similar.

Theorem 3: If one angle of a triangle is equal to the one angle of another triangle and the sides including these angles are proportional, then, the two triangles are said to be similar.

Now let us understand how to solve each question step-by-step.

Class 10 Maths Exercise 6.3 Solutions

Exercise 6.3 Class 10 Question 1

Out of the 16 questions in Maths Class 10 Chapter 6 Exercise 6.3, the first one is based on the similarity of pairs of triangles. A set of pairs of triangles are given, and you need to identify which ones are similar and also mention the criterion based on which you define them as similar triangles. You also need to give the correct notation for describing similar triangles.

Exercise 6.3 Class 10 Question 2

In the second question of Class 10 Maths Chapter 6 Exercise 6.3, students need to apply the relationship of adjacent angles and similarity of triangles to find out angles of two similar triangles which are formed by the intersection of two straight lines which are intersecting two parallel lines.

Exercise 6.3 Class 10 Question 3

For this question of NCERT Class 10 Maths ch 6 ex 6.3, students are given a trapezium ABCD, and they need to prove that for the point of intersection O, the proportion of the lengths of the diagonals AO/OC = OB/OD. You need to apply the theorem for alternate interior angles and vertically opposite angles to solve this problem.

Exercise 6.3 Class 10 Question 4

In this question, a triangle is given, which contains two triangles inside of it. By applying the theorem that if the sides of a triangle are proportional, then their corresponding angles are equal, you can prove that the two interior triangles are similar.

Exercise 6.3 Class 10 Question 5

This is a straightforward question where two angles of two triangles are the same (AA criterion) hence students can prove that the given triangles are similar.

Exercise 6.3 Class 10 Question 6

This question of Class 10 Maths Exercise 6.3 involves applying the knowledge of congruent triangles and CPCT (corresponding parts of congruent triangles) to prove that the given triangles are similar.

Exercise 6.3 Class 10 Question 7

In this question of Triangles Class, 10 Exercise 6.3, the altitudes of two triangles are given, and you need to prove that the triangles made by the intersection of the altitudes are similar. You will be using the concept of vertically opposite angles and common angles to solve this question.

Exercise 6.3 Class 10 Question 8

This question has a parallelogram ABCD, and a straight line from point B intersects DC at F and extends till point E. The point D is extended to meet E, which forms triangle AEB. Using the property of opposite angles of a parallelogram and the concept of alternate interiors angels students need to prove that △ABE and △CFB are similar.

Exercise 6.3 Class 10 Question 9

This is again a simple question where two right triangles are given which have a common side, and you need to prove that both are similar triangles which can be easily proved with a common angle, corresponding sides of similar triangles, and alternate angle concept.

Exercise 6.3 Class 10 Question 10

Class 10 Maths Exercise 6.3 question 10 involves the angle bisector theorem and alternate angle similarity criterion to prove that two triangles formed by bisecting an angle are similar.

Exercise 6.3 Class 10 Question 11

In this problem, an isosceles triangle is given, and one of its sides is extended, and a perpendicular is dropped from the extended point to the opposite side of the triangle. Students need to prove that the triangle formed by extension and the triangle formed by dropping a perpendicular to the base of the original triangle are similar. The properties of the isosceles triangle and alternate angle criterion are used to solve this problem.

Exercise 6.3 Class 10 Question 12

There are two triangles ABC and PQR in this problem, and it is given that the medians of the two triangles and the adjacent sides are proportional. You need to prove triangles ABC and PQR are similar triangles. The median divides the opposite side, using this concept along with SSS (for congruent triangles) and corresponding angles of similar triangles, this problem can be solved.

Exercise 6.3 Class 10 Question 13

In a triangle, ABC, a straight line AD on the side BC is drawn, and ∠ADC is equal to ∠BAC. One needs to show that CA² = CB.CD. This is done using concepts of common angle, alternate angle, and corresponding angles of similar triangles rules.

Exercise 6.3 Class 10 Question 14

This question is similar to question 12 where the sides of two triangles and their median are proportional, and you need to prove that the two triangles are similar. This question is solved by extending the median of both the triangles by equal length and then using rules of a parallelogram, SSS, SAS, and corresponding angles of similar triangles criterion to prove that the triangles are similar.

Exercise 6.3 Class 10 Question 15

This is a question on a vertical pole and its shadow along with the shadow of a tower. Given the dimensions of the pole’s height and shadow length, you need to find the height of the tower. Using the AAA (angle-angle-angle) theorem of similarity one can deduce the tower’s height.

Exercise 6.3 Class 10 Question 16

Two similar triangles are given, and students need to prove that the ratio of sides of the triangles and their medians are equal. Using the property of the median and corresponding angle similarity criterion, one can solve this problem.

Key Features of NCERT Solutions for Class 10 Maths Exercise 6.3

All the problems in the Ex 6.3 Class 10 NCERT Solutions are prepared meticulously by the expert team of Vedantu, and the key takeaways of these solutions are:

• The solutions are available in downloadable PDF format, which can also be printed out for a quick revision.

• The expert team of Vedantu has based all the answers on the CBSE curriculum; hence you can expect to score high marks in maths.

• The solutions are free of cost.

NCERT Solutions for Class 10 Maths Chapter 6 All Other Exercises